Premed · Premed · Calculus 1
Lecture 18: Antiderivatives
Calculus I — Differential Calculus
Learning Objectives
By the end of this lecture, students will be able to:
- Define an antiderivative and the indefinite integral
- Apply basic antidifferentiation rules (power rule, trig, exponential, logarithmic)
- Use the "+C" convention and explain why it is necessary
- Solve initial value problems using antiderivatives
- Relate antiderivatives to motion (position from velocity, velocity from acceleration)
Lecture Content
I. Definition of Antiderivative
A function F is an antiderivative of f on an interval I if F'(x) = f(x) for all x in I. For example, F(x) = x^3 is an antiderivative of f(x) = 3x^2, because d/dx[x^3] = 3x^2. Antiderivatives are not unique: if F is an antiderivative of f, then so is F(x) + C for any constant C.
A theorem guarantees that if F is any one antiderivative of f on an interval, then every antiderivative of f has the form F(x) + C. This follows from the Mean Value Theorem: if F'(x) = G'(x) for all x, then F(x) - G(x) must be a constant.
II. Indefinite Integral Notation
The indefinite integral of f(x) is written as integral of f(x) dx = F(x) + C, where F'(x) = f(x) and C is the constant of integration. The integral sign is an elongated S (hinting at the connection to summation that will appear with definite integrals). The function f(x) is called the integrand, and dx indicates the variable of integration. It is essential to always include +C for indefinite integrals -- omitting it is one of the most common errors in calculus.
III. Basic Antidifferentiation Rules
The Power Rule in reverse states that the integral of x^n dx = x^{n+1}/(n+1) + C for n not equal to -1. For instance, the integral of x^4 dx = x^5/5 + C, the integral of x^{-3} dx = -1/(2x^2) + C, and the integral of sqrt(x) dx = (2/3)x^{3/2} + C.
The special case n = -1 requires separate treatment: the integral of 1/x dx = ln|x| + C. The constant multiple rule says the integral of c f(x) dx = c integral of f(x) dx, and the sum/difference rule allows term-by-term integration.
IV. Antiderivatives of Common Functions
The trigonometric antiderivatives are: integral of cos x dx = sin x + C, integral of sin x dx = -cos x + C, integral of sec^2 x dx = tan x + C, integral of csc^2 x dx = -cot x + C, integral of sec x tan x dx = sec x + C, and integral of csc x cot x dx = -csc x + C.
The exponential antiderivatives are: integral of e^x dx = e^x + C, and integral of a^x dx = a^x / ln a + C (for a > 0, a not equal to 1).
Two inverse-trig-related antiderivatives are particularly important: integral of 1/sqrt(1 - x^2) dx = arcsin x + C, and integral of 1/(1 + x^2) dx = arctan x + C.
<image>A two-column reference table of basic antiderivatives. Left column shows the integrand f(x); right column shows the antiderivative F(x) + C. Includes: power rule (x^n for general n and n = -1), trig functions (sin, cos, sec^2, csc^2, sectan, csccot), exponentials (e^x, a^x), and inverse trig (1/sqrt(1-x^2), 1/(1+x^2)). Title: "Table of Basic Antiderivatives."</image>
V. Antidifferentiation by Rewriting
Many integrands must be rewritten before a basic rule applies. Expanding products is one strategy: the integral of (x + 1)^2 dx = integral of (x^2 + 2x + 1) dx = x^3/3 + x^2 + x + C. Dividing each term is another: the integral of (x^3 + 2x)/x dx = integral of (x^2 + 2) dx = x^3/3 + 2x + C. Rewriting radicals and fractions as powers makes the power rule applicable: the integral of 1/x^4 dx = integral of x^{-4} dx = x^{-3}/(-3) + C. Using trig identities can also simplify integrands: the integral of tan^2 x dx = integral of (sec^2 x - 1) dx = tan x - x + C.
VI. Initial Value Problems
An initial value problem (IVP) provides a differential equation together with an initial condition, which determines the constant C. The general process is: first, find the general antiderivative (with +C); second, substitute the initial condition to solve for C; third, write the particular solution.
For example, given f'(x) = 6x - 2 and f(1) = 5, the general antiderivative is f(x) = 3x^2 - 2x + C. The initial condition gives f(1) = 3 - 2 + C = 5, so C = 4, and the particular solution is f(x) = 3x^2 - 2x + 4.
VII. Motion Applications
Antiderivatives connect acceleration to velocity and velocity to position. Given acceleration a(t), the velocity is v(t) = integral of a(t) dt + C_1, where the initial velocity v(0) determines C_1. The position is then s(t) = integral of v(t) dt + C_2, where the initial position s(0) determines C_2.
For free fall, a(t) = -g = -9.8 m/s^2, and integrating twice with initial conditions v(0) = v_0 and s(0) = s_0 gives v(t) = -9.8t + v_0 and s(t) = -4.9t^2 + v_0 t + s_0. As another example, if a particle has acceleration a(t) = 12t - 6, initial velocity v(0) = 4, and initial position s(0) = 1, then v(t) = 6t^2 - 6t + 4 and s(t) = 2t^3 - 3t^2 + 4t + 1.
<image>A three-panel vertical stack showing the relationships between acceleration, velocity, and position. Panel A (top): a(t) = constant (horizontal line), with an arrow labeled "integrate" pointing down to Panel B. Panel B (middle): v(t) = linear function, with an arrow labeled "integrate" pointing down to Panel C. Panel C (bottom): s(t) = quadratic function (parabola). Reverse arrows labeled "differentiate" point upward. The initial conditions v(0) and s(0) are shown determining the integration constants. Title: "From acceleration to position via antiderivatives."</image>
